Problem: You have found the following ages (in years) of all 6 sloths at your local zoo: $ 4,\enspace 19,\enspace 14,\enspace 16,\enspace 6,\enspace 12$ What is the average age of the sloths at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 sloths at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{4 + 19 + 14 + 16 + 6 + 12}{{6}} = {11.8\text{ years old}} $ Find the squared deviations from the mean for each sloth. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $4$ years $-7.8$ years $60.84$ years $^2$ $19$ years $7.2$ years $51.84$ years $^2$ $14$ years $2.2$ years $4.84$ years $^2$ $16$ years $4.2$ years $17.64$ years $^2$ $6$ years $-5.8$ years $33.64$ years $^2$ $12$ years $0.2$ years $0.04$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{60.84} + {51.84} + {4.84} + {17.64} + {33.64} + {0.04}} {{6}} $ $ {\sigma^2} = \dfrac{{168.84}}{{6}} = {28.14\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{28.14\text{ years}^2}} = {5.3\text{ years}} $ The average sloth at the zoo is 11.8 years old. There is a standard deviation of 5.3 years.